In exothermic reactions, there is a release heat and the replacement of weak bonds with stronger ones.
We are given the number of moles:
O = 1.36 mol
H = 4.10 mol
C = 2.05 mol
To get the empirical formula, first divide everything by
the smallest number of moles = 1.36 mol. So that:
O = 1 mol
H = 3 mol
C = 1.5 mol
Next step is to multiply everything by a number such that
all will be a whole number. In this case, multiply by 2 to get a whole number
for C, so that:
O = 2 mol
H = 6 mol
C = 3 mol
Therefore the empirical formula of the compound is:
C3H6O2
Answer:
1. 0.45 mole
2. 49.95g
Explanation:
The following were obtained from the question:
Volume of solution = 300mL = 300/1000 = 0.3L
Molarity = 1.5 M
Mole of CaCl2 =?
1. We can obtain the mole of the solute as follow:
Molarity = mole of solute /Volume of solution
1.5 = mole of solute/0.3
Mole of solute = 1.5 x 0.3
Mole of solute = 0.45 mole
2. The grams in 0.45 mole of CaCl2 can be obtained as follow:
Molar Mass of CaCl2 = 40 + (35.5 x 2) = 40 + 71 = 111g/mol
Mole of CaCl2 = 0.45 mole
Mass of CaCl2 =?
Mass = number of mole x molar Mass
Mass of CaCl2 = 0.45 x 111
Mass of CaCl2 = 49.95g
Answer:
95.2 - 40.8 = 54.4 g of oxygen
number of moles = mass (g)/ Mr
no. of moles of carbon = 40.8/12 = 3.4
no. of moles of oxygen = 3.4
divide both by smallest value which is 3.4 and you’ll get 1 mole of carbon and 1 mole of oxygen therefore the empirical formula is CO
Explanation:
hope this helps :)
